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4. Amazon executives believe that at least 70% of customers would return a product 2 days after it arrives at their home. A sample of 500 customers found 68% returned the product they purchased prior to the third day. Given the executives' estimate, what would be the probability of a sample result with 68% or fewer returns prior to the third day

Sagot :

Using the normal distribution and the central limit theorem, it is found that there is a 0.1635 = 16.35% probability of a sample result with 68% or fewer returns prior to the third day.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportions has mean [tex]\mu = p[/tex] and standard error [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]

In this problem:

  • Sample of 500 customers, hence [tex]n = 500[/tex].
  • Amazon believes that the proportion is of 70%, hence [tex]p = 0.7[/tex]

The mean and the standard error are given by:

[tex]\mu = p = 0.7[/tex]

[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{500}} = 0.0205[/tex]

The probability is the p-value of Z when X = 0.68, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.68 - 0.7}{0.0205}[/tex]

[tex]Z = -0.98[/tex]

[tex]Z = -0.98[/tex] has a p-value of 0.1635.

0.1635 = 16.35% probability of a sample result with 68% or fewer returns prior to the third day.

A similar problem is given at https://brainly.com/question/25735688